Using the digits 1, 2, 3, 4, 5, 6, 7, a number of 4 different digits is formed. Find. Match each item in Colum

NCERT Exemplar · Q63

1, 2, 3, 4, 5, 6, 7

(a) how many numbers are formed?;

(b) how many numbers are exactly divisible by

2

25

(d) how many of these are exactly divisble by

4

C_2

(i)

840

;

(ii)

200

;

(iii)

360

;

(iv)

40

Odisha ChseShort· 3mImportance★★★★★est

99% · 129/130 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Permutations Without Repetition

Permutations Without Repetition – The Idea of Arranging Things

Imagine you have three different books on a shelf: a Physics book, a Chemistry book, and a Maths book. How many different ways can you arrange them in a row?

You could try listing them out:

Physics, Chemistry, Maths
Physics, Maths, Chemistry
Chemistry, Physics, Maths
Chemistry, Maths, Physics
Maths, Physics, Chemistry
Maths, Chemistry, Physics

That's 6 arrangements. Notice that each arrangement uses all three books exactly once — no book is repeated, and no book is left out. This is the core idea: permutations without repetition count the number of ways to arrange a set of distinct objects in order, using each object exactly once.

Why "Without Repetition"?

The phrase "without repetition" means that once you place an object in a position, you cannot use it again. In our book example, once you put the Physics book in the first slot, you cannot put it in the second or third slot. Each object appears exactly once in the arrangement.

This is different from "permutations with repetition" (like creating 3-letter codes from the letters A, B, C where you can reuse letters — e.g., AAA, AAB, etc.). Here, no repeats allowed.

The Counting Logic – Why Multiply?

Let's build the arrangement step by step for 3 distinct books:

First position: You have 3 choices (any of the 3 books).
Second position: After placing the first book, only 2 books remain — so 2 choices.
Third position: Only 1 book is left — so 1 choice.

Total arrangements = $3 \times 2 \times 1 = 6$ .

This product $3 \times 2 \times 1$ is called 3 factorial, written as $3!$ .

P(n) = n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1

For $n$ distinct objects, the number of permutations (arrangements in order) is $n!$ .

What If You Only Arrange Some of Them?

Suppose you have 5 different books, but you only want to arrange 3 of them on a shelf. How many ways?

First position: 5 choices
Second position: 4 choices
Third position: 3 choices

Total = $5 \times 4 \times 3 = 60$ .

This is a permutation of 5 objects taken 3 at a time, written as $P(5,3)$ or ${}^5P_3$ .

P(n, r) = \frac{n!}{(n-r)!} = n \times (n-1) \times \cdots \times (n-r+1)

Here $n$ is the total number of distinct objects, and $r$ is how many you are arranging. The formula works because:

Numerator $n!$ counts all arrangements of all $n$ objects.
Denominator $(n-r)!$ removes the arrangements of the $n-r$ objects you are not using.

Key Points to Remember

Important

Order matters — swapping two objects gives a different permutation.
No repetition — each object is used at most once.
For arranging all $n$ objects: $n!$
For arranging r out of $n$ objects: $\frac{n!}{(n-r)!}$

Common Mistake to Avoid …

Start your 14-day free trial to unlock the full solution →

We form 4-digit numbers from $\{1,2,3,4,5,6\}$ without repetition. Count all arrangements, then apply divisibility rules: divisible by 2 needs even last digit; by 25 needs last two digits 25 or 75; by 4 needs last two digits divisible by 4. (a)→(i) 840, (b)→(iii) 360, (c)→(iv) 40, (d)→(ii) 200.

The core idea is permutations without repetition: we're selecting and arranging 4 distinct digits from 7 available digits. The total count is straightforward, but divisibility constraints force us to think strategically—fix the constrained positions first, then count what remains.

(a) How many numbers are formed?

We need to choose 4 digits from 7 and arrange them. The first position has 7 choices, the second has 6 remaining, the third has 5, and the fourth has 4.

P(7, 4) = 7 \times 6 \times 5 \times 4 = 840

P(n, r) = \frac{n!}{(n-r)!}

So (a) matches (i) 840.

(b) How many numbers are exactly divisible by 2?

A number is divisible by 2 if and only if its last digit is even. From our set $\{1,2,3,4,5,6,7\}$ , the even digits are $\{2, 4, 6\}$ .

Strategy: Fix the last digit first, then arrange the remaining three positions.

Choose the last digit: 3 choices (2, 4, or 6).
Fill the remaining three positions: We have 6 digits left, and we need to arrange 3 of them.

P(6, 3) = 6 \times 5 \times 4 = 120

Total: $3 \times 120 = 360$

So (b) matches (iii) 360.

(c) How many numbers are exactly divisible by 25?

A number is divisible by 25 if its last two digits form a number divisible by 25. The two-digit multiples of 25 are: 25, 50, 75.

From our digits $\{1,2,3,4,5,6,7\}$ :

25: Both 2 and 5 are available. ✓
50: We don't have 0. ✗
75: Both 7 and 5 are available. ✓

So the last two digits must be either 25 or 75.

Fix the last two digits: 2 choices (25 or 75).
Fill the first two positions: We've used 2 digits, so 5 remain. We arrange 2 of them.

P(5, 2) = 5 \times 4 = 20

Total: $2 \times 20 = 40$

So (c) matches (iv) 40.

(d) How many numbers are exactly divisible by 4?

A number is divisible by 4 if its last two digits form a number divisible by 4.

We need to identify all two-digit combinations from $\{1,2,3,4,5,6,7\}$ (with distinct digits) that are divisible by 4.

| Last two digits | Value | Divisible by 4? | …

Start your 14-day free trial to unlock the full solution →

←Previous questionQ62 — There are 10 professors and 20 lecturers out of whom a committee of 2 …→Next questionQ64 — How many words (with or without dictionary meaning) can be made from t…

Q.Using the digits 1,2,3,4,5,6,71, 2, 3, 4, 5, 6, 71,2,3,4,5,6,7, a number of 444 different digits is formed. Find. Match each item in Column C1C_1C1​ with its correct answer in Column C2C_2C2​. C1C_1C1​:

(a) How many numbers are formed?

(b) How many numbers are exactly divisible by 2?

(c) How many numbers are exactly divisible by 25?

(d) How many numbers are exactly divisible by 4?

🎓Unlock everything free for 14 days

Q.Using the digits $1, 2, 3, 4, 5, 6, 7$ , a number of $4$ different digits is formed. Find. Match each item in Column $C_1$ with its correct answer in Column $C_2$ . $C_1$ :

Unlock everything free for 14 days